Proving that $\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\frac{x-1}{x-2}, x \in \mathbb{R}_{ >2}$

Question

While doing numerical experiments with series involving binomial coefficients at Mathematica I stumbled up on $$\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\frac{x-1}{x-2}, \quad x \in \mathbb{R}_{\ >2}.$$ The question is: How to prove it by hand?

Answer 1

${x+k-1 \choose k}^{-1}=\dfrac{(x-1)!k!}{(x+k-1)!}$

Introduce $\Gamma$ function:

$\dfrac{\Gamma(x)\Gamma(k+1)}{\Gamma(x+k)}=\dfrac{(x-1)\Gamma(x-1)\Gamma(k+1)}{\Gamma(x+k)}=(x-1)\beta(x-1, k+1)$

where $\beta((x-1, k+1)=\int\limits_0^1t^{x-2}(1-t)^kdt$

Put it back into the sum:

$(x-1)\int\limits_0^1t^{x-2}\sum\limits_{k=0}^\infty (1-t)^kdt=(x-1)\int\limits_0^1t^{x-3}dt=\dfrac{x-1}{x-2}$

Answer 2

Note that $${n \choose k}^{-1}=(n+1)\int_{0}^{1} t^k (1-t)^{n-k} dt$$ Then $$S=\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\sum_{k=0}^{\infty} (k+x) \int_{0}^{1} t^k (1-t)^{x-1} dt$$ $$\implies S=\int_{0}^{1} (1-t)^{k-1} \sum_{k=0}^{\infty} (k+x) t^k dt= \int_{0}^{1} (1-t)^{x-1}\left( \frac{t}{(1-t)^2}+\frac{x}{1-t} \right)dt$$ $$\implies S=\int_{0}^{1} [ (x-1)(1-t)^{x-2}+(1-t)^{x-3}]~ dt=\frac{x-1}{x-2}, x >2.$$

Answer 3

Another Gamma Approach $$ \begin{align} \sum_{k=0}^\infty\binom{x+k-1}{k}^{-1} &=1+\sum_{k=1}^\infty\frac{k\,\Gamma(k)\,\Gamma(x)}{\Gamma(x+k)}\tag1\\ &=1+\sum_{k=1}^\infty k\int_0^1t^{k-1}(1-t)^{x-1}\,\mathrm{d}t\tag2\\ &=1+\int_0^1\frac{(1-t)^{x-1}}{(1-t)^2}\,\mathrm{d}t\tag3\\[6pt] &=1+\frac1{x-2}\tag4\\[6pt] &=\frac{x-1}{x-2}\tag5 \end{align} $$ Explanation:
$(1)$: pull the $k=0$ term out of the sum
$\phantom{\text{(1):}}$ write in terms of $\Gamma$
$(2)$: use the Beta function
$(3)$: $\sum_{k=1}^\infty kt^{k-1}=(1-t)^{-2}$
$(4)$: integrate
$(5)$: add